加入三个表mysql php，但没有回音

Question

What did I do wrong. It echo nothing. The expect output is apple and pear with val good and 0. First, I will get all fruit with price 4 on table_1, then I join table_2, exclude any fruit mark=4, exclude banana, then I join table_3, get the val if factor match 56, so apple match, and pear factor did not match, so return 0. But my code echo nothing, didnot see error, but just echo nothing.

/*
table_1
fruit  price
apple   4
apple   5
banana  4
banana  5
pear    4

table_2
fruit   mark
apple    5
apple    6
banana   4
banana   7
pear     6

table_3
fruit   factor  val
apple    56     good
apple    60      OK
banana   89     good
banana   90     good
pear     56      bad
*/
$pri=4;
$sql = $wpdb->get_results( $wpdb->prepare("
                 SELECT 
                 b.fruit,c.val
                 FROM table_1 a,
                 LEFT JOIN table_2 b
                  ON b.fruit=a.fruit and b.mark != '4'
                 LEFT JOIN table_3 c
                  ON c.fruit=a.fruit AND c.factor = '56'
                WHERE a.price=%d
                 ",$pri));
print_r(sql);

</div>

Answer 1

Just Remove (,) after FROM (SELECT DISTINCT fruit FROM table_1 WHERE price=%d) a,

 /*table_1
 fruit  price
 apple   4
 apple   5
 banana  4
 banana  5
 pear    4

table_2
fruit   mark
apple    5
apple    6
banana   4
banana   7
pear     6

table_3
fruit   factor  val
apple    56     good
apple    60      OK
banana   89     good
banana   90     good
pear     56      bad
*/
$pri=4;
$sql = $wpdb->get_results( $wpdb->prepare("
                 SELECT 
                 b.fruit,c.val
                 FROM (SELECT DISTINCT fruit FROM table_1 WHERE price=%d) a
                 LEFT JOIN table_2 b
                  ON b.fruit=a.fruit and b.mark != '4'
                 LEFT JOIN table_3 c
                  ON c.fruit=a.fruit AND c.factor = '56'
                 ",$pri));
print_r(sql);

And the output is:

fruit   val     
apple   good
apple   good
pear    bad
banana  NULL